Một lăng trụ tứ giác đều có cạnh đáy a, đường chéo của lăng trụ tạo với  cạnh bên và với đường chéo đáy hai góc có tổng bằng 1350. Xác định đường cao lăng trụ.

Kí hiệu góc giữa  đường chéo D1B và đường chéo đáy DBy và góc giữa D1BBCx, theo giả thiết x+y=1350
Ta có BC=cosx.BD1=aBD=cosy.BD1=a2}(1)
Do (1) ta có: cosxcosy=12cosycos(1350y)=2
Giải phương trình: cosy=2cos(1350y)
                  cos(1350y)=cos(450+y)=22(cosysiny)
Nên cosy+siny=cosysiny=2cosytany=2
Đường cao DD1=BDtany=a2tany=2a2

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